Mathematics · Quantitative Aptitude

Equations and Roots

69 Questions

Equations and roots questions cover finding real solutions to algebraic, trigonometric, and polynomial equations. They are fundamental for advanced mathematics sections. Practicing these problems ensures quick recognition of underlying patterns and calculation shortcuts.

polynomial equationssystem of equationstrigonometric equationsabsolute value equations

Equations and Roots Questions

Multiple choice general knowledge math & puzzles

The equation 1/a + 1/|a| = 0 has

  1. an infinite number of solutions

  2. no solutions

  3. 2 solutions only

  4. 3 solutions only
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation $1/a + 1/|a| = 0$ implies $1/|a| = -1/a$. This is true for all negative values of $a$. If $a < 0$, then $|a| = -a$, so $1/(-a) = -1/a$, which is an identity. Thus, there are infinite solutions (all $a < 0$).

Multiple choice general knowledge science & technology

The three solutions of the equation f(x) = 0 are -2, 0, and 3. Therefore, the three solutions of the equation f(x - 2) = 0 are
    • 4, -2, and 1

  2. -2, 0 and 3

  3. 4, 2, and 5

  4. 0, 2 and 5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If f(x) = 0 has solutions at -2, 0, and 3, then for f(x - 2) = 0, we substitute x - 2 for x. This means x - 2 = -2, 0, or 3, giving x = 0, 2, and 5. The transformation shifts all solutions 2 units to the right.

Multiple choice general knowledge science & technology

What is Drake's Equation?

  1. N = N*fp*ne*fl*fi*fc*fL

  2. x^n = y^n + z^n , n>2

  3. (a+b)/a = a/b

  4. (a*b)/(c*d) = 1/a/b/c/d
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Drake's Equation is N = N* × fp × ne × fl × fi × fc × fL, used to estimate the number of active, communicative extraterrestrial civilizations in the Milky Way galaxy. It was formulated by astrophysicist Frank Drake in 1961. Option B is Fermat's Last Theorem (proven by Andrew Wiles). Option C is related to the Golden Ratio. Option D is not a standard equation.

Multiple choice maths lines graphs of linear equations equations of lines parallel to the x-axis and y-axis graph of linear equations in two variables

If $4x+3y=120$, find how many non-negative integer solutions are possible?

  1. $1$
  2. $11$

  3. Infinite

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We can write the equation in another form $4(x-3)+3(y+4)=120$
Now, we make a table

x 30 27 24 21 18 15 12 9 6 3 0
y 0 4 8 12 16 20 24 28 32 36 40

 We observe a patter that $x$ reduces by $3$ and $y$ increases by $4$. But both $x$ and $y$ cannot be negative or $0$. So, the value of $x=0$ and $y=0$ is ruled out. Hence, nine such positive integer solutions are possible.
Alternately, we can write the given equation as $4(x+3)+3(y-4)=120$. We get the same values of $x$ and $y$

Multiple choice business maths limits and continuity of a function graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

The solution set of the equation $(x+1)^2+[x-1]^2=(x-1)^2+[x+1]^2$ where $[x]$ and $(x)$ are the greatest integer and nearest integer to $x$, is 

  1. $ x\in R$
  2. $x\in N$
  3. $x\in I$
  4. $x\in Q$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation is (x+1)^2 + [x-1]^2 = (x-1)^2 + [x+1]^2. Rearranging gives (x+1)^2 - [x+1]^2 = (x-1)^2 - [x-1]^2. Let f(t) = t^2 - [t]^2. We need f(x+1) = f(x-1). Since f(t) = {t}^2 + 2{t}[t], this holds when x is an integer.

Multiple choice decimal representation of rational numbers rational and irrational numbers maths

Number of solutions of the equation $[2x]-3{2x}=1$ is?

(where $[\cdot]$ and ${\cdot }$ denote greatest integer an fractional part function respectively).

  1. $1$
  2. $2$
  3. $3$
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solving [2x] - 3{2x} = 1. Let y = 2x. Then [y] - 3{y} = 1. Since {y} = y - [y], we have [y] - 3(y - [y]) = 1, which simplifies to 4[y] - 3y = 1, or y = (4[y] - 1)/3. Testing integer values for [y] gives valid solutions for y in the range [n, n+1).

Multiple choice maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

The number of solution of $z^2 + \bar{z} = 0$ is

  1. $5$
  2. $4$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $z=x+iy$.
Now,
$z^2+\overline{z}=0$
or, $x^2-y^2+2ixy+(x-iy)=0$
or, $(x^2-y^2+x)+i(2xy-y)=0$
Now comparing the real and imaginary part both sides we get,
$x^2-y^2+x=0$.....(1) and $2xy-y=0$.....(2).
From (2) we get, $x=\dfrac{1}{2}$ or $y=0$.
Now $x=\dfrac{1}{2}$ gives from (1) we get, $y=\pm \dfrac{\sqrt{3}}{2}$.
And $y=0$ gives from (1) we get, $x=0, 1$.
So the solution s are $(0,0), (1,0), \left(\dfrac{1}{2},\pm \dfrac{\sqrt{3}}{2}\right)$.
So we have $4$ solutions.
Multiple choice maths set concepts finite and infinite sets types of sets set language

If the system of equation $x+2y-3z=1$, $(p+2)z=3$, $(2p+1)y+z=2$ has infinite number of solutions, then the value of p is not equal to.

  1. $-2$
  2. $-\displaystyle\frac{1}{2}$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$x+2y-3z=1$
$(p+2)z=3$
$(2p+1)y+z=2$
let $p=5$, $s\in R/ \left\{ -2, 1/2\right\}$
$\therefore z=\dfrac{3}{s+2}$
$\Rightarrow y=\left(2-\dfrac{3}{s+2}\right)\dfrac{1}{(2s+1)}\Rightarrow \dfrac{2s+1}{(s+2)(2s+1)}-\dfrac{1}{s+2}$
$[As\ 2s+1\neq 0]$
$\therefore x=3z+1-2y$
$=\dfrac{9}{s+2}+1-\dfrac{2}{s+2}=\dfrac{7}{s+2}+1$
$\therefore$ solutions $(x, y, z)=\left(\dfrac{7}{s+2}+1, \dfrac{1}{s+2}, \dfrac{3}{s+2}\right)$
is an infinite set,
$\therefore p$ cannot be equal to $-2$ or $1/2$
Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

The number of real solution of the equation $(\dfrac{9}{10})^x=-3+x-x^2$ is-

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $f\left( x \right) ={ \left( \cfrac { 9 }{ 10 }  \right)  }^{ x }$
$g\left( x \right) =-3+x-{ x }^{ 2 }$
For $x<0$
$f\left( x \right) >1$
$g\left( x \right) -1=-{ x }^{ 2 }+x-4$
$\triangle =1-4\left( -1 \right) \left( -4 \right) $
$=-15<0$
and co efficient of ${ x }^{ 2 }=-1<0$
$\therefore g\left( x \right) -1<0$
$\therefore g\left( x \right) <1$
$\therefore f\left( x \right) -g\left( x \right) $ has no solution if $x<0$
For $x>0$
$f\left( x \right) <1$
$-3+x-{ x }^{ 2 }=f\left( x \right) $
$-{ x }^{ 2 }+x-3-f\left( x \right) =0$
$\triangle =1-4\left( -1 \right) \left( -3-f\left( x \right)  \right) $
$=1-12+4f\left( x \right) $
$\triangle =-11+4f\left( x \right) $
$f\left( x \right)<0$
$4f\left( x \right)-11<-11$
$\therefore \triangle <0$
$f\left( x \right)=g\left( x \right)$ has no real solution for $x>0$
At $x=0$
$f\left( x \right)=1,g\left( x \right)=-3$
$\therefore f\left( x \right)=g\left( x \right)$ has no real solution $\forall x\epsilon R$
Multiple choice reciprocal equations theory of equations maths

The solution set of the equation 
$x^{2/3} + x^{1/3} = 2 $ is

  1. $\{-8, 1\}$
  2. $\{8, 1\}$
  3. $\{1, -1\}$
  4. $\{2, -2}\$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let ${ x }^{ \cfrac { 1 }{ 3 }  }=t\ { t }^{ 2 }+t-2=0\ { t }^{ 2 }+2t-t-2=0\ t(t+2)-(t+2)=0\ (t-1)(t+2)=0\ t=1\ x^{ \cfrac { 1 }{ 3 }  }=1\ x=1\ t=-2\ x^{ \cfrac { 1 }{ 3 }  }=-2\ x=-8\ x=\left( -8,1 \right)$

Multiple choice reciprocal equations theory of equations maths

The equation $\sin^{-1}x-3\sin^{-1}a=0$ has real solutions for x if?

  1. $a \in R$
  2. $a \in [-1, 1]$
  3. $a \in \left[0, \dfrac{1}{2}\right]$
  4. $a \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$
Reveal answer Fill a bubble to check yourself
A Correct answer